Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm actually an engineering student so I'm not too good with probability and was hoping someone may be able to help with the following:

So I have a ratio of discrete random variables. I want to be able to know when I can get away with approximating its expected value as a ratio of expected values. For example, for the continuous case I came across the following formulation from a Taylor series expansion

$\text{E}[x/y]=\text{E}[x]/ \text{E}[y]- \text{Cov}[x ,y]/ \text{E}[y]^2 + \text{E}[x] \text{Var}[y]/ \text{E}[y]^3$

Does this apply for discrete random variables? if not, what is its analogous expression?

thank you! hadi

share|cite|improve this question
How did you derive the formula? Thanks! – qed Sep 5 '13 at 15:38

A simple, general, good approximation purely in terms of expectations does not seem achievable. For example, let $\epsilon$ be a very small positive number, and let $X$ and $Y$ be independent, each taking on the value $\epsilon$ with probability $1/2$, and $1$ with probability $1/2$. Then $E(X/Y)$ is very large, but the various moments are all quite reasonable.

Possibly some reasonable approximations are possible if all values of $Y$ are well away from $0$.

share|cite|improve this answer

That last term doesn't make sense to me. Let $E[x] = \mu_x$ and $E[y] = \mu_y$. From the Taylor series of $x/t$ around $t=\mu_y$, $x/y = x/\mu_y - x(y -\mu_y)/\mu_y^2 + x (y - \mu_y)^2/\mu_y^3 + \ldots $ resulting in $E[x/y] = \mu_x/\mu_y - {\rm Cov}[x,y]/\mu_y^2 + E[x (y - \mu_y)^2]/\mu_y^3 + \ldots$, but $E[x (y - \mu_y)^2] \ne \mu_x {\rm Var}[y]$ unless $x$ and $(y - \mu_y)^2$ are uncorrelated.

In any case, it has nothing to do with discrete vs. continuous: it is just as valid for discrete random variables as it is for continuous ones. The main point is that to have a good approximation (apart from that last problematic term) you'll want $y - E[y]$ to be (with high probability) small compared to $|E[y]|$. You especially need to avoid $y$ near 0, which could make $E[x/y]$ be undefined.

share|cite|improve this answer

First of all, $X$ and $Y$ should be independent. Now the problem is that in general $E[1/y] \neq 1/E[y]$. However, this is a reasonable approximation when $y$ is concentrated around some "large" value.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.